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abdelhamid belaid
Joined: 13 Aug 2009 Posts: 170

Posted: Fri May 03, 2013 10:59 pm Post subject: 


this is a formula without "atan(tan())" function for the dodecahedron (I did the same before in fact):
Code:  max(
max(
sin(1.07498)*sqrt(x*x+y*y)*cos(abs(abs(abs(abs(atan2(y,x))pi/5)pi/5)pi/5)pi/5)+cos(1.07498)*z
,
sin(1.07498)*sqrt(x*x+y*y)*cos(abs(abs(abs(atan2(abs(y),x)2*pi/5)pi/5)pi/5)pi/5)cos(1.07498)*z )
,
abs(z) )3
[x]: 4 , 4
[y]: 4 , 4
[z]: 4 , 4 
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abdelhamid belaid
Joined: 13 Aug 2009 Posts: 170

Posted: Sat May 04, 2013 12:04 am Post subject: 


hexagonal grid torus :
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abdelhamid belaid
Joined: 13 Aug 2009 Posts: 170

Posted: Sat May 04, 2013 7:51 am Post subject: 


I'd like to add, Furan meant plane1^n+ ... +plane5^n+z^na^n , it's the same principle when making cube x^n+y^n+z^na^n , I had used it when made the letters (A to Z) for the first time, that's great and here I recommend to write (plane1/a)^n + ... +(plane5/a)^n+(z/a)^n1:
Code:  ((sin(1.07498)*sqrt(x*x+y*y)*cos(atan2(y,x))+cos(1.07498)*z )/3)^30+
((sin(1.07498)*sqrt(x*x+y*y)*cos(atan2(y,x)2*pi/5)+cos(1.07498)*z )/3)^30+
((sin(1.07498)*sqrt(x*x+y*y)*cos(atan2(y,x)4*pi/5)+cos(1.07498)*z )/3)^30+
((sin(1.07498)*sqrt(x*x+y*y)*cos(atan2(y,x)+2*pi/5)+cos(1.07498)*z )/3)^30+
((sin(1.07498)*sqrt(x*x+y*y)*cos(atan2(y,x)+4*pi/5)+cos(1.07498)*z )/3)^30+
(z/3)^30
1
[x]: 4 , 4
[y]: 4 , 4
[z]: 3.2 , 3.2 
the problem here is the hard and long work you must do to find every plane formula which no one likes, also other shapes have more faces!
in fact (Denis) I didn't mean just a writing without min and max, that's very easy (we know max(a,b)=(a+b+abs(ab))/2 and min(a,b)=(a+babs(ab))/2) ), I meant another issue (how could us create many similar faces starting from one), we need this to save time and also for other cases, like this:
another formula for the dodecahedron "sqrt(aabs(plane1))*...*sqrt(aabs(plane6)) = 0" or for round edges "sqrt(aabs(plane1))*...*sqrt(aabs(plane6)) c^2= 0"
Code:  sqrt(3abs(sin(1.07498)*sqrt(x*x+y*y)*cos(atan2(y,x))+cos(1.07498)*z ) )*
sqrt(3abs(sin(1.07498)*sqrt(x*x+y*y)*cos(atan2(y,x)2*pi/5)+cos(1.07498)*z ))*
sqrt(3abs(sin(1.07498)*sqrt(x*x+y*y)*cos(atan2(y,x)4*pi/5)+cos(1.07498)*z ) )*
sqrt(3abs(sin(1.07498)*sqrt(x*x+y*y)*cos(atan2(y,x)+2*pi/5)+cos(1.07498)*z ) )*
sqrt(3abs(sin(1.07498)*sqrt(x*x+y*y)*cos(atan2(y,x)+4*pi/5)+cos(1.07498)*z ) )*
sqrt(3abs(z))
0.2
x]: 4 , 4
[y]: 4 , 4
[z]: 3.1 , 3.1 
_________________ My YouTube channel
Last edited by abdelhamid belaid on Mon May 20, 2013 4:54 pm; edited 7 times in total 

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abdelhamid belaid
Joined: 13 Aug 2009 Posts: 170

Posted: Sat May 04, 2013 2:17 pm Post subject: 


also for the dodecahedron we would write "(aabs(plane1))*...*(aabs(plane6)) +c= 0" :
Code:  (3abs(sin(1.07498)*sqrt(x*x+y*y)*cos(atan2(y,x))+cos(1.07498)*z ) )*
(3abs(sin(1.07498)*sqrt(x*x+y*y)*cos(atan2(y,x)2*pi/5)+cos(1.07498)*z ))*
(3abs(sin(1.07498)*sqrt(x*x+y*y)*cos(atan2(y,x)4*pi/5)+cos(1.07498)*z ) )*
(3abs(sin(1.07498)*sqrt(x*x+y*y)*cos(atan2(y,x)+2*pi/5)+cos(1.07498)*z ) )*
(3abs(sin(1.07498)*sqrt(x*x+y*y)*cos(atan2(y,x)+4*pi/5)+cos(1.07498)*z ) )*
(3abs(z))^2
0.03
[x]: 4 , 4
[y]: 4 , 4
[z]: 3.1 , 3.1 
_________________ My YouTube channel
Last edited by abdelhamid belaid on Sat May 04, 2013 4:11 pm; edited 1 time in total 

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abdelhamid belaid
Joined: 13 Aug 2009 Posts: 170

Posted: Sat May 04, 2013 2:53 pm Post subject: 


I have a question, what's more important: describing shape mathematically such that you know every thing about or creating just shape close to another one??, for me I must know every detail about the shape I imagine, every point position, every angle, every length ....., just round edge is not enough to say I know what I do I think!, you have picture doesn't mean you know it!, there is big difference between math and art, for certain shapes we can make them with clay, that's faster !!! _________________ My YouTube channel 

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inode
Joined: 27 Jan 2007 Posts: 127 Location: Austria

Posted: Tue May 07, 2013 5:16 am Post subject: truncated Dodecahedron 


Did you know that if you replace the constant of your formula to 16.666 you'll get a truncated dodecahedron...
Code:  Name: PG_TruncDodeca1
/* Isosurface: PG_TruncDodeca1 20130506
looks like a truncated dodecahedron */
F(): sqrt(3abs(sin(1.07498)*sqrt(x*x+y*y)*cos(atan2(y,x) )+cos(1.07498)*z))
* sqrt(3abs(sin(1.07498)*sqrt(x*x+y*y)*cos(atan2(y,x)2*pi/5)+cos(1.07498)*z))
* sqrt(3abs(sin(1.07498)*sqrt(x*x+y*y)*cos(atan2(y,x)4*pi/5)+cos(1.07498)*z))
* sqrt(3abs(sin(1.07498)*sqrt(x*x+y*y)*cos(atan2(y,x)+2*pi/5)+cos(1.07498)*z))
* sqrt(3abs(sin(1.07498)*sqrt(x*x+y*y)*cos(atan2(y,x)+4*pi/5)+cos(1.07498)*z))
* sqrt(3abs(z))
16.666
[x]: 1 , 1
[y]: 1 , 1
[z]: 1 , 1
; 
Last edited by inode on Thu May 16, 2013 6:46 am; edited 1 time in total 

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abdelhamid belaid
Joined: 13 Aug 2009 Posts: 170

Posted: Tue May 07, 2013 12:42 pm Post subject: 


I did not Gerd, I just tried low numbers to get round edges, thanks for the nice shape _________________ My YouTube channel 

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abdelhamid belaid
Joined: 13 Aug 2009 Posts: 170

Posted: Wed May 08, 2013 5:58 pm Post subject: 


hello,
Taha in Fri Sep 07, 2007 at 5:03 am had wrote a post "mirrored isosurface":
http://k3dsurf.s4.bizhat.com/k3dsurfftopic59.html
Quote:  Hi all,
You can use this formula to obtain a mirrored view of an isosurface F() :
Quote:  G(x,y,z) = F(x,y,z)* F(x  2*P(x,y,z)*a/R,
y  2*P(x,y,z)*b/R,
z  2*P(x,y,z)*c/R)
Where :
P(x,y,z) = ax+by+cz+d (Equation of the Plan)
R = sqrt(a^2 + b^2 + c^2) 
Maybe for future use in minimal surface?
Enjoy 
I would like just to remind you this great small math simulation and basic 3d math processing, it's very late I know, I like this, with such basic processing (functions) we can do much (displace, scale, twist, mirrored isosurface, multiplicity, patterned cube making, patterned spher making, ......................... ), maybe we should gather a box of tools (functions) to create 3d math drawing "programing" language , maybe in the future we would build great 3d drawing softwares (easier and faster) _________________ My YouTube channel 

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abdelhamid belaid
Joined: 13 Aug 2009 Posts: 170


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denisc
Joined: 24 Apr 2013 Posts: 92

Posted: Sun May 12, 2013 6:29 am Post subject: 


hello all
a view of my bad isocaedron
[/img] 

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abdelhamid belaid
Joined: 13 Aug 2009 Posts: 170

Posted: Sun May 12, 2013 10:30 am Post subject: 


Denis, I think you have to try more Grid resolution or try other limits _________________ My YouTube channel 

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denisc
Joined: 24 Apr 2013 Posts: 92

Posted: Sun May 12, 2013 7:09 pm Post subject: 


hello
not exactly
it's a bad formula
(you can see the formula, in my precedent post
more resolution
also another formula for icosa, but the wall of ico are flat, not right.
z^65*(x^2+y^2)*z^4+5*(x^2+y^2)^2*z^22*(x^410*x^2*y^2+5*y^4)*x*z(x^2+y^2+z^2)^3(x^2+y^2+z^2)+1


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abdelhamid belaid
Joined: 13 Aug 2009 Posts: 170

Posted: Mon May 13, 2013 6:19 pm Post subject: 


the second one is fantastic Denis
Quote:  also another formula for icosa, but the wall of ico are flat, not right.
z^65*(x^2+y^2)*z^4+5*(x^2+y^2)^2*z^22*(x^410*x^2*y^2+5*y^4)*x*z(x^2+y^2+z^2)^3(x^2+y^2+z^2)+1

how beautiful to see works for others, there are many users on this forum but they are very silent!!, every work whatever has its special taste, thanks Denis for sharing _________________ My YouTube channel
Last edited by abdelhamid belaid on Tue May 14, 2013 12:53 pm; edited 2 times in total 

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denisc
Joined: 24 Apr 2013 Posts: 92

Posted: Mon May 13, 2013 9:35 pm Post subject: 


thanks mr belaid
This is a backup, old enough, I do not remembered me.
(I recall me the source of the formula, perhaps on this forum, or on the web
so I'm a little confused by the fact that I no longer recall what just me and what comes from another )
I just found a little pressure, the same formula for the dodeca in my files
and the formula/
z^65*(x^2+y^2)*z^4+5*(x^2+y^2)^2*z^22*(x^410*x^2*y^2+5*y^4)*x*z+(x^2+y^2+z^2)^3(x^2+y^2+z^2)^2+(x^2+y^2+z^2)1
So I looked so alone, 6 months ago, the formula of dodeca and icosa.
I do not happen to find straight edges and
I am discouraged and I forgot.
cheers
denis 

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inode
Joined: 27 Jan 2007 Posts: 127 Location: Austria

Posted: Wed May 15, 2013 5:33 am Post subject: Icosahedron 


Hi all
Expanding the formula of a dodecahedron from abdelhamid belaid the corresponding formula of an icosahedron is like the following...
Code:  Name: PG_Icosahedron1
/* Isosurface: PG_Icosahedron1 20130514 Gerd Platl
looks like a rounded icosahedron */
F(): 0.1
 ((sin(0.666)*sqrt(x*x+y*y)*cos(atan2(y,x) )+cos(0.666)*z )/3)^30
 ((sin(0.666)*sqrt(x*x+y*y)*cos(atan2(y,x)2*pi/5)+cos(0.666)*z )/3)^30
 ((sin(0.666)*sqrt(x*x+y*y)*cos(atan2(y,x)4*pi/5)+cos(0.666)*z )/3)^30
 ((sin(0.666)*sqrt(x*x+y*y)*cos(atan2(y,x)+2*pi/5)+cos(0.666)*z )/3)^30
 ((sin(0.666)*sqrt(x*x+y*y)*cos(atan2(y,x)+4*pi/5)+cos(0.666)*z )/3)^30
 ((sin(1.37)*sqrt(x*x+y*y)*cos(atan2(y,x) )+cos(1.37)*z )/3)^30
 ((sin(1.37)*sqrt(x*x+y*y)*cos(atan2(y,x)2*pi/5)+cos(1.37)*z )/3)^30
 ((sin(1.37)*sqrt(x*x+y*y)*cos(atan2(y,x)4*pi/5)+cos(1.37)*z )/3)^30
 ((sin(1.37)*sqrt(x*x+y*y)*cos(atan2(y,x)+2*pi/5)+cos(1.37)*z )/3)^30
 ((sin(1.37)*sqrt(x*x+y*y)*cos(atan2(y,x)+4*pi/5)+cos(1.37)*z )/3)^30
[x]: 3.1 , 3.1
[y]: 3.1 , 3.1
[z]: 3.31 , 3.31
; 
If you take a look at http://en.wikipedia.org/wiki/Icosahedron to the picture with 3 interconnected orthogonal golden rectangles where the edges are connected by red lines you'll see the 1st angle can be calculated by sin(alpha)=1.0/goldenRatio.
Rearranging give us the 1st angle:
alpha[rad] = arcSin(1/1.618)=0.666.
But the 2nd angle value is only set approximately. If anybody know how to get the 2nd angle value of the other 10 planes (at the moment set to 1.37), please let me know.
Gerd 

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